## 9.28 Review

In this section we give a quick review of what has transpired above.

Let $K/k$ be a field extension. Let $\alpha \in K$. Then we have the following possibilities:

The element $\alpha $ is transcendental over $k$.

The element $\alpha $ is algebraic over $k$. Denote $P(T) \in k[T]$ its *minimal polynomial*. This is a monic polynomial $P(T) = T^ d + a_1 T^{d - 1} + \ldots + a_ d$ with coefficients in $k$. It is irreducible and $P(\alpha ) = 0$. These properties uniquely determine $P$, and the integer $d$ is called the *degree of $\alpha $ over $k$*. There are two subcases:

The polynomial $\text{d}P/\text{d}T$ is not identically zero. This is equivalent to the condition that $P(T) = \prod _{i = 1, \ldots , d} (T - \alpha _ i)$ for pairwise distinct elements $\alpha _1, \ldots , \alpha _ d$ in the algebraic closure of $k$. In this case we say that $\alpha $ is *separable* over $k$.

The $\text{d}P/\text{d}T$ is identically zero. In this case the characteristic $p$ of $k$ is $ > 0$, and $P$ is actually a polynomial in $T^ p$. Clearly there exists a largest power $q = p^ e$ such that $P$ is a polynomial in $T^ q$. Then the element $\alpha ^ q$ is separable over $k$.

Definition 9.28.1. Algebraic field extensions.

A field extension $K/k$ is called *algebraic* if every element of $K$ is algebraic over $k$.

An algebraic extension $k'/k$ is called *separable* if every $\alpha \in k'$ is separable over $k$.

An algebraic extension $k'/k$ is called *purely inseparable* if the characteristic of $k$ is $p > 0$ and for every element $\alpha \in k'$ there exists a power $q$ of $p$ such that $\alpha ^ q \in k$.

An algebraic extension $k'/k$ is called *normal* if for every $\alpha \in k'$ the minimal polynomial $P(T) \in k[T]$ of $\alpha $ over $k$ splits completely into linear factors over $k'$.

An algebraic extension $k'/k$ is called *Galois* if it is separable and normal.

The following lemma does not seem to fit anywhere else.

Lemma 9.28.2. Let $K$ be a field of characteristic $p > 0$. Let $L/K$ be a separable algebraic extension. Let $\alpha \in L$.

If the coefficients of the minimal polynomial of $\alpha $ over $K$ are $p$th powers in $K$ then $\alpha $ is a $p$th power in $L$.

More generally, if $P \in K[T]$ is a polynomial such that (a) $\alpha $ is a root of $P$, (b) $P$ has pairwise distinct roots in an algebraic closure, and (c) all coefficients of $P$ are $p$th powers, then $\alpha $ is a $p$th power in $L$.

**Proof.**
It follows from the definitions that (2) implies (1). Assume $P$ is as in (2). Write $P(T) = \sum \nolimits _{i = 0}^ d a_ i T^{d - i}$ and $a_ i = b_ i^ p$. The polynomial $Q(T) = \sum \nolimits _{i = 0}^ d b_ i T^{d - i}$ has distinct roots in an algebraic closure as well, because the roots of $Q$ are the $p$th roots of the roots of $P$. If $\alpha $ is not a $p$th power, then $T^ p - \alpha $ is an irreducible polynomial over $L$ (Lemma 9.14.2). Moreover $Q$ and $T^ p - \alpha $ have a root in common in an algebraic closure $\overline{L}$. Thus $Q$ and $T^ p - \alpha $ are not relatively prime, which implies $T^ p - \alpha | Q$ in $L[T]$. This contradicts the fact that the roots of $Q$ are pairwise distinct.
$\square$

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